# Compact Metric Spaces and Inverse Functions ... Apostol, Example After Theorem 4.29 ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...

I am focused on Chapter 4: Limits and Continuity ... ...

I need help in order to fully understand the example given after Theorem 4.29 ... ...

Theorem 4.29 (including its proof) and the following example read as follows:  In the Example above we read the following:

" ... ... However, $$\displaystyle f^{ -1 }$$ is not continuous at the point $$\displaystyle f(0)$$. For example, if $$\displaystyle x_n = 1 - 1/n$$, the sequence $$\displaystyle \{ f(x_n) \}$$ converges to $$\displaystyle f(0)$$ but $$\displaystyle \{ x_n \}$$ does not converge in $$\displaystyle S$$. ... ... "

My question is as follows:

Can someone please explain exactly how/why ... the sequence $$\displaystyle \{ f(x_n) \}$$ converges to $$\displaystyle f(0)$$ but $$\displaystyle \{ x_n \}$$ does not converge in $$\displaystyle S$$ ... ... implies that $$\displaystyle f^{ -1 }$$ is not continuous at the point $$\displaystyle f(0)$$ ... ... ?

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My thoughts ...

I think that the relevant theorem regarding answering my question is Apostol, Theorem 4.16 which reads as follows: If we let $$\displaystyle t_n \in f(S)$$ be such that $$\displaystyle t_n = f(x_n)$$ ... so the sequence $$\displaystyle \{ t_n \} = \{ f(x_n) \}$$ is in the domain of $$\displaystyle f^{ -1 }$$ ...

Then ... sequence $$\displaystyle \{ t_n \} = \[ f(x_n) \}$$ converges to $$\displaystyle f(0) = t_0$$ say ...

Then following Theorem 4.16 above ... for $$\displaystyle f^{ -1 }$$ to be continuous we need $$\displaystyle \{ f^{ -1 } (t_n) \} = \{ f^{ -1 } ( f(x_n) ) \} = \{ x_n \}$$ to converge in $$\displaystyle S$$ ... but it does not do so ...

(mind you ... I'm not sure how to prove it doesn't converge in $$\displaystyle S$$ ...)

Is that correct?

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Hope that someone can help ...

Peter

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#### castor28

##### Well-known member
MHB Math Scholar
Hi Peter ,

Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$.

Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps discontinuously from $(1-\varepsilon)$ to $0$ in $S$ whenever $P$ passes through $f(0)=1$.

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#### Peter

##### Well-known member
MHB Site Helper
Hi Peter ,

Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$.

Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps discontinuously from $(1-\varepsilon)$ to $0$ in $S$ whenever $P$ passes through $f(0)=1$.

Thanks castor28